There are many commonly asked questions regarding C programming language. Below are some collected such question-answer examples. The questions are usually related with 32-bit system, Turbo C IDE in windows or GCC under Linux environment [not always].
For more such examples, click C_Q&A label.
Predict the output or error(s) for the following:
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main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");break;
case 2:printf("two");break;
case 3: printf("three");break;
}
}
Answer:
Three
Explanation:
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
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main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation:
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
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main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error: Type mismatch in redeclaration of function display
Explanation:
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
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main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
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#define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
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main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
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#include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M"); So we get the output 77 :: "M" (Ascii is 77).
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#include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values.
*q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
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#include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
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…till next post, bye-bye & take care.
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